GCSE Revision: Indices

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Positive powers

To write repeated addition we use multiplication:
$\displaystyle 2 + 2 + 2 + 2 = 2 \times 4$
... this saves lots of time writing out additions of the same thing over and over.

Similarly we might want to find a shorthand for writing out repeated mulitplication. For this we use exponents (sometimes called powers or indices):
$\displaystyle 2 \times 2 \times 2 \times 2 = 2^4$

In this example, the number 2 is the base and 4 is the exponent or index (plural: 'indices'). The whole of the 24 should be read as "Two to the four", or "Two to the power of four."

In general:
$\displaystyle x^a = \underbrace {x \times x \times x \ldots \times x}_{\mbox{$a$ lots of $x$ multiplied together}}$

Basic index laws

There are three rules you must learn and be able to use accurately:

1.	$\displaystyle x^a \times x^b = x^{a + b}$	examples: $\displaystyle 7^3 \times 7^2 = (7 \times 7 \times 7) \times (7 \times 7) = 7^5$
2.
3.

• x^a times x^b = x^{a+b}

For example,

5^2 times 5^3 = (5 times 5) times (5 times 5 times 5) = 5^5

.

• frac{x^a}{x^b} = x^{a-b}

For example,

frac{3^4}{3^2} = frac{3 times 3 times 3 times 3}{3 times 3} = frac{3 times 3}{1} = 3^2

.

• (x^a)^b = x^{ab}

For example,

(5^2)^3 = 5^2 times 5^2 times 5^2 = 5^6

.
===Common pitfalls===
Now, this isn't too difficult, but people often make two mistakes. The first one is to assume that because

x^a times x^b

and

frac{x^a}{x^b}

can be simplified easily, so can

x^a times y^b

and

frac{x^a}{y^b}

, where

x ne y

. Wrong. They can only be simplified if

y

is a power of

x

or vice versa. So, for example:

3^4 times 2^4

can't be simplified.

5^2 times 25^2 = 5^2 times (5^2)^2 = 5^2 times 5^4 = 5^6

can be simplified.

The second one is to assume that

x^{a^b} = (x^a)^b

. They look equal, but they really aren't. For example:

(5^3)^2 = 5^3 times 5^3 = 5^6

5^{3^2} = 5^9

See what I mean?
==Zero and negative powers==
Now, some bright spark noticed that the division identity above only worked for

b < a

, and decided to define exponentiation for zero and negative exponents. Obviously, the key consideration when doing something like that is to make sure that zero and negative exponents behave as much like positive exponents as possible - that is, make them obey all three of the above identities. He eventually settled on the following definition for

x^0

:

x^0 = 1

for all x.

I know, not an intuitive definition, right? But it's the only sensible definition under which all three identities hold:

x^a times x^0 = x^a times 1 = x^a = x^{a+0}

frac{x^a}{x^0} = frac{x^a}{1} = x^a = x^{a-0}

(x^a)^0 = 1 = x^0 = x^{a times 0}

(x^0)^a = 1^a = 1 = x^0 = x^{0 times a}

As for negative integers, it turns out from the division identity that the only sensible definition is:

x^{-a} = frac{1}{x^a}

Again, all three identities hold:

x^a times x^{-b} = frac{x^a}{x^b} = x^{a-b}

frac{x^a}{x^{-b}} = x^a times x^b = x^{a+b}

(x^a)^{-b} = frac{1}{(x^a)^b} = frac{1}{x^{ab}} = x^{-ab}

(x^{-a})^b = (frac{1}{x^a})^b = frac{1}{x^{ab}} = x^{-ab}

This also renders the identity

frac{x^a}{x^b} = x^{b-a}

redundant, since dividing by

x^b

is the same thing as multiplying by

x^{-b}

. For this reason, sometimes

x^a times x^b = x^{a+b}

is known as the law of exponentiation.
==Rational powers==
But of course, this wasn't enough! Someone decided that integer powers were so good they might as well generalise to rational powers - in other words, fractions. As before, the key thing was to make sure that all the old identities worked. Assuming they do, then:

(x^{frac{1}{a}})^a = x^{atimesfrac{1}{a}} = x

So the obvious choice for

x^{frac{1}{a}}

was:

x^{frac{1}{a}} = sqrt[a]{x}

Then we have:

x^{frac{a}{b}} = (x^{a})^{frac{1}{b}} = sqrt[b]{x^a}

It turns out that both our identities hold, so we can take this as our definition. You might want to skip the proofs, though - they're a little messy, and they're not on the syllabus. Here they are anyway for completeness:

begin{eqnarray[*]}(x^{frac{a}{b}})^{frac{c}{d}} & = & sqrt[b]{x^a}^{frac{c}{d}} \& = & sqrt[d]{(sqrt[b]{x^a})^c} \& = & sqrt[bd]{sqrt[d]{(sqrt[b]{x^a})^c}^{bd}} \& = & sqrt[bd]{((sqrt[b]{x^a})^c)^b} \& = & sqrt[bd]{(sqrt[b]{x^a})^{bc}} \& = & sqrt[bd]{(x^a)^c} \& = & sqrt[bd]{x^{ac}} \& = & x^{frac{ac}{bd}} \end{eqnarray[*]}

begin{eqnarray[*]}x^{frac{a}{b}} times x^{frac{c}{d}} & = & (x^{frac{a}{b}})^1 times (x^{frac{c}{d}})^1 \& = & (x^{frac{a}{b}})^{frac{d}{d}} times (x^{frac{c}{d}})^{frac{b}{b}} \& = & x^{frac{ad}{bd}} times x^{frac{bc}{bd}} \& = & sqrt[bd]{x^{ad}} times sqrt[bd]{x^{bc}} \& = & sqrt[bd]{(sqrt[bd]{x^{ad}} times sqrt[bd]{x^{bc}})^{bd}} \end{eqnarray[*]}

begin{eqnarray[*]}& = & sqrt[bd]{(sqrt[bd]{x^{ad}})^{bd} times (sqrt[bd]{x^{bc}})^{bd}} \& = & sqrt[bd]{x^{ad} times x^{bc}} \& = & sqrt[bd]{x^{ad + bc}} \& = & x^{frac{ad + bc}{bd}} \& = & x^{frac{a}{b} + frac{c}{d}} \end{eqnarray[*]}

==Complex powers==
It's possible to define exponentiation for irrational exponents like

sqrt{2}

, and complex exponents like

2 + 3sqrt{-1}

, but this is beyond the scope of A Level.
==Summary of Identities==

x^a = underbrace{x times x times ldots times x}_{text{a total x's in the product}}

x^0 = 1

x^{-a} = frac{1}{x^a}

x^{frac{1}{a}} = sqrt[a]{x}

x^{a} times x^{b} = x^{a+b}

(x^a)^b = x^{ab}

==Examples==

2^{-2} = frac{1}{4}

8^{frac{1}{3}} = 2

32^{frac{2}{5}} = 4

2^{2^{2^2}} = 2^{2^4} = 2^{16} = 65536